How do you use implicit differentiation to find dy/dx given #y=(x+y)^2#?
2 Answers
I got
#(dy)/(dx) = (2x + 2y)/(1 - 2x - 2y)#
Since
#d/(dx)[y(x)] = d/(dx)[(x+y)^2]#
#(dy)/(dx) = 2(x+y)^(1) * stackrel("Chain Rule")overbrace(d/(dx)[x+y])#
#= 2(x+y)(1 + (dy)/(dx))#
#= (2x + 2y)(1 + (dy)/(dx))#
#= 2x + 2x (dy)/(dx) + 2y + 2y(dy)/(dx)#
Now isolate the similar terms.
#(dy)/(dx) - 2x(dy)/(dx) - 2y(dy)/(dx) = 2x + 2y#
#(dy)/(dx)[1 - 2x - 2y] = 2x + 2y#
#=> color(blue)((dy)/(dx) = (2x + 2y)/(1 - 2x - 2y))#
Explanation:
Here is an alternative answer.
Expand inside the parentheses:
Hopefully this helps!